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Problem solution · Java

Minimum Sum of Values by Dividing Array

Minimum Sum of Values by Dividing Array: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
42 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Minimum Sum of Values by Dividing Array, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 42 lines of Java from the credited upstream file 3117.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Sum of Values by Dividing Array · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumValueSum(int[] nums, int[] andValues) {    Map<Integer, Integer>[][] mem = new Map[nums.length][andValues.length];    Arrays.stream(mem).forEach(A -> Arrays.setAll(A, j -> new HashMap<>()));    final int ans = minimumValueSum(nums, andValues, 0, 0, FULL_MASK, mem);    return ans == INF ? -1 : ans;  }   private static final int INF = 1_000_000_000;  private static final int FULL_MASK = (1 << 17) - 1;   // Returns the minimum value sum of nums[i..n) and andValues[j..m), where  // `mask` is the running value of the current subarray.  private int minimumValueSum(int[] nums, int[] andValues, int i, int j, int mask,                              Map<Integer, Integer>[][] mem) {    if (i == nums.length && j == andValues.length)      return 0;    if (i == nums.length || j == andValues.length)      return INF;    if (mem[i][j].containsKey(mask))      return mem[i][j].get(mask);    mask &= nums[i];    if (mask < andValues[j]) {      mem[i][j].put(mask, INF);      return INF;    }    if (mask == andValues[j]) {      // 1. Keep going.      // 2. End the subarray here and pick nums[i], then fresh start.      final int res =          Math.min(minimumValueSum(nums, andValues, i + 1, j, mask, mem),                   nums[i] + minimumValueSum(nums, andValues, i + 1, j + 1, FULL_MASK, mem));      mem[i][j].put(mask, res);      return res;    }    // Keep going.    final int res = minimumValueSum(nums, andValues, i + 1, j, mask, mem);    mem[i][j].put(mask, res);    return res;  }}

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